KC Sinha Mathematics Solution Class 11 Chapter 9 गणितीय आगमन का सिद्धांत (Principle of mathematical induction) Exercise 9.1

Exercise 9.1

Question 1

$1+2+3+\ldots+n=\frac{n(n+1)}{2}$
Sol :
STEP-1
Let P(n):$1+2+3+\ldots+n=\frac{n(n+1)}{2}$

putting n=1

L.H.S=n=1

R.H.S=$=\frac{n(n+1)}{2}=\frac{1(1+1)}{2}=\frac{1 \times 2}{2}=1$

L.H.S=R.H.S

which means P(1) is true

STEP-2
Let P(k) is true

P(k): $1+2+3+\ldots+n=\frac{k(k+1)}{2}$..(i)

then prove that P(k+1) is also true

P(k+1): $1+2+3+...+k+(k+1)=\frac{(k+1)(k+2)}{2}$..(ii)

adding both sides (k+1) in (i)

P(k+1): $1+2+3+\ldots+k+(k+1)=\frac{k(k+1)}{2}+(k+1)$

$=(k+1)\left[\frac{k}{2}+1\right]$

$=(k+1)\left(\frac{k+2}{2}\right)$

$=\frac{(k+1)(k+2)}{2}$

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 2

1+3+5+..+(2n-1)=n²

Sol :
Let P(n): 1+3+5+..+(2n-1)=n²
STEP-1
when n=1

L.H.S=2(1)-1
=2-1=1

R.H.S=1²=1

Thus, P(1) is true. 

STEP-2

Let P(k) be true. Then,

P(k): 1+3+5+..+(2k-1)=k²..(i)

then prove that P(k+1) is true

$P(k+1): 1+3+5+\ldots (2 k-1)+(2 k+1)=(k+1)^{2}$..(ii)

adding both sides (2k+1) in (i)

P(k+1): 1+3+5...+(2k-1)+(2k+1)=k²+2k+1

=k²+2.k.1+1²

=(k+1)²

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 3

$1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}$
Sol :
Let P(n): $1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}$

Step-1

when n=1

L.H.S=1²=1

R.H.S=$=\frac{1(1+1)(2 \times 1+1)}{6}$

$=\frac{1 \times 2 \times 3}{6}=1$

L.H.S=R.H.S

Step-2
Let P(k) be true. Then,

$P(k) : 1^{2}+2^{2}+3^{2}+\ldots \ldots+k^{2}=\frac{k(k+1)(2 k+1)}{6}$..(i)

then prove that P(k+1) is true

$P(k+1): 1^{2}+2^{2}+3^{2}+\cdots-k^{2}+(k+1)^{2}=\frac{(k+1)(k+2)(2 k+3)}{6}$..(ii)

adding (k+1)² in (i) both sides

$1^{2}+2^{2}+3^{2}--+k^{2}+(k+1)^{2}=k\frac{(k+1)(2 k+1)}{6}+(k+1)^2$

$=(k+1)\left[\frac{k(2 k+1)}{6}+(k+1)\right]$

$=(k+1)\left[\frac{k(2 k+1)+6(k+1)}{6}\right]$

$=\frac{(k+1)\left(2 k^{2}+k+6 k+6\right)}{6}$

$=\frac{(k+1)\left(2 k^{2}+7 k+6\right)}{6}$

$=\dfrac{(k+1)\left[2 k^{2}+4 k+3 k+6\right]}{6}$

$=\frac{(k+1)[2 k(k+2)+3(k+2)]}{6}$

$=\dfrac{(k+1)(k+2)(2 k+3)}{6}$

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 5

$a+a r+a r^{2}+\ldots+a r^{n-1}=\frac{a\left(1-r^{n}\right)}{1-r}$
Sol :
Let $a+a r+a r^{2}+\ldots+a r^{n-1}=\frac{a\left(1-r^{n}\right)}{1-r}$

Step-1
when n=1

L.H.S=ar^1-1=ar⁰=a×1=a

R.H.S=$=\frac{a\left(1-r\right)}{1-r}$
$=\frac{a(1-r)}{1-r}=a$

P(1) is true

Step-2

Let P(k) be true. Then,

$P(k): a+a r+a r^{2}+\ldots+a r^{k-1}=\frac{a\left(1-r^{k}\right)}{1-r}$..(i)

then prove that P(k+1) is true
$P(k+1): a+a r+a r^{2}+\dots a r^{k-1}+ar^{k}=\frac{a\left(1-r^{k+1}\right)}{1-r}$..(ii)

adding both sides ar^k in (i) we get
$a+a r+a r^{2}+\dots+a r^{k-1}+a r^{k}=\frac{a\left(1-r^{k}\right)}{1-r}+a r^{k}$

$=\frac{a-a r^{k}+a r^{k}-a r^{k+1}}{1-r}$

$=\frac{a\left(1-r^{k+1}\right)}{1-r}$

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 6

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}$
Sol :
Let $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}$

Step-1
when n=1

L.H.S$=\frac{1}{2^{1}}=\frac{1}{2}$

R.H.S$=1-\frac{1}{2^{1}}=1-\frac{1}{2}$
$=\frac{2-1}{2}=\frac{1}{2}$

P(1) is true

Step-2
Let P(k) be true. Then,

$P(k): \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\quad---\frac{1}{2^{k}}=1-\frac{1}{2^{k}}$..(i)

then prove that P(k+1) is true

$P(k+1): \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+---\frac{1}{2 k}+\frac{1}{2^{k+1}}=1-\frac{1}{2^{k+1}}$..(ii)

adding both side $\frac{1}{2^{k+1}}$ in (i)

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+----+\frac{1}{2^{k}}+\frac{1}{2^{k+1}}=1-\frac{1}{2^{k}}+\frac{1}{2^{k+1}}$

$=1-\frac{1}{2^{k}}+\frac{1}{2^{1} \cdot 2^{k}}$

$=1-\frac{1}{2^{k}}\left(1-\frac{1}{2}\right)$

$=1-\frac{1}{2^{k}}\left(\frac{2-1}{2}\right)$

$=1-\frac{1}{2^{k}} \times \frac{1}{2}$

$=1-\frac{1}{2^{k+1}}$

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 8

$1 \cdot 2+2 \cdot 2^{2}+3 \cdot 2^{3}+\ldots+n \cdot 2^{n}=(n-1) 2^{n+1}+2$

Sol :

P(n): $1 \cdot 2+2 \cdot 2^{2}+3 \cdot 2^{3}+\ldots+n \cdot 2^{n}=(n-1) 2^{n+1}+2$

Step-1

when n=1

l.H.S=$1 \cdot 2^{1}=1 \times 2=2$

R.H.S=$=(1-1) 2^{1+1}+2=0 \times 2^{2}+2=2$

L.H.S=R.H.S

P(1) is true

Step-2

Let P(k) be true. Then,

$P(k): 1.2+2.2^{2}+3.2^{3}+\ldots+k \cdot 2^{k}=(k-1)2^{k+1}+2$..(i)

then prove that P(k+1) is true

$P(k+1): \quad 1 \cdot 2+2 \cdot 2^{2}+3 \cdot 2^{3}+\ldots \ldots \quad+(k+1) 2^{k+1}=k \cdot 2^{k+2}+2$..(ii)

adding both sides $(k+1) \cdot 2^{k+1}$ in (i)

$1 \cdot 2+2 \cdot 2^{2}+3 \cdot 2^{3}+\ldots+k \cdot 2^{k}+(k+1) \cdot 2^{k+1}=(k-1) \cdot 2^{k+1}+2+(k+1) \cdot 2^{k+1}$

$=2^{k+1}(k-1+k+1)+2$

$=2^{1} k^{1} \cdot 2^{k+1}+2$

$=k \cdot 2^{k+2}+2$

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 19

$1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+n}=\frac{2 n}{n+1}$
Sol :
P(n): $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+n}=\frac{2 n}{n+1}$

Step-1
when n=1

L.H.S$=\frac{1}{1+2+3+\ldots+n}=\frac{1}{\frac{n(n+1)}{2}}$

$=\frac{2}{n(n+1)}$
$=\frac{2}{1(1+1)}=\frac{2}{1 \times 2}=1$

R.H.S$=\frac{2 n}{n+1}=\frac{2 \times 1}{1+1}=\frac{2}{2}=1$

P(1) is true

Step-2
Let P(k) be true. Then,

$P(k):1+\frac{1}{1+2}+\frac{1}{1+2+3}+---\frac{1}{1+2+3+\ldots+k}=\frac{2 k}{k+1}$..(i)

prove that P(k+1) is true

$P(k+1): \quad 1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots(k+1)=\frac{2(k+1)}{k+2}$..(ii)

adding both sides $\frac{1}{1+2+3+\dots+(k+1)}$ in (i)

$1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots \cdot \cdot+\frac{1}{1+2+3 \cdots+k}+\frac{1}{1+2+3+\ldots(k+1)}=\frac{2 k}{k+1}+\frac{1}{1+2+3+\ldots+(k+1)}$

$=\frac{2 k}{k+1}+\frac{1}{\frac{k+1}{2}[1+k+1]}$

$=\frac{2 k}{k+1}+\frac{1}{(k+1)(k+2)}$

$=\frac{2 k}{k+1}+\frac{2}{(k+1)(k+2)}$

$=\frac{2}{k+1}\left[k+\frac{1}{k+2}\right]$

$=\frac{2}{k+1}\left[\frac{k^{2}+2 k+1}{k+2}\right]$

$=\frac{2}{k+1} \times \frac{(k+1)^{2}}{k+2}$

$=\frac{2(k+1)}{k+2}$

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 20

$\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \dots\left(1+\frac{1}{n}\right)=(n+1)$
Sol :
P(n): $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \dots\left(1+\frac{1}{n}\right)=(n+1)$
Step-1
when n=1

L.H.S $=1+\frac{1}{2}=1+1=2$

R.H.S=1+1=2

L.H.S=R.H.S

P(1) is true

Step-2
Let P(k) be true. Then,

$P(k):(1++)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \cdots\left(1+\frac{1}{k}\right)=k+1$..(i)

Prove that P(k+1) is true

$\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots-\left(1+\frac{1}{k+1}\right)=k+2$..(ii)

multiplying both sides by $\left(1+\frac{1}{k+1}\right)$ in (i)

$\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots \quad\left(1+\frac{1}{k}\right) \cdot\left(1+\frac{1}{k+1}\right)=(k+1)\left(1+\frac{1}{k+1}\right)$

$=(k+1)\left(\frac{k+1+1}{k+1}\right)=k+2$

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 22

$(\cos \theta+i \sin \theta)^{n}=\cos n \theta+i \sin n \theta$
Sol :
P(n): $(\cos \theta+i \sin \theta)^{n}=\cos n \theta+i \sin n \theta$

Step-1

when n=1

L.H.S $=(\cos \theta+i \sin \theta)^{1}=\cos \theta+i \sin \theta$

R.H.S$=\cos (1) \theta+i \sin (1) \theta=\cos \theta+i \sin \theta$

L.H.S=R.H.S

P(1) is true

Step-2
Let P(k) be true. Then,
$P(k):(\cos \theta+i \sin \theta)^{k}=\cos k \theta+i \sin k \theta$..(i)

prove that P(k+1) is true

$P(k+1): ( \cos \theta+i \sin \theta)^{k+1}=\cos (k+1) \theta+i \sin (k+1) \theta$..(ii)

multiplying both sides by $\left(\cos \theta+i \sin \theta\right)$ in (i)

$(\cos \theta+i \sin \theta)^{k} \cdot(\cos \theta+i \sin \theta)^{1}=(\cos k \theta+i \sin k \theta)(\cos \theta+1 \sin \theta)$

$(\cos \theta+i \sin \theta)^{k+1}=\cos k \theta \cos \theta+i \cos \theta \sin \theta+i\sin k \theta \cos \theta + i^2 \sin k \theta \sin \theta$

=(coskθcosθ-sinkθsinθ)+i(sinkθcosθ+coskθsinθ)

=cos(kθ+θ)+isin(kθ+θ)

=cos(k+1)θ+isin(k+1)θ

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 23

$\cos \theta \cdot \cos 2 \theta \cdot \cos 4 \theta \ldots \cos 2^{n-1} \theta=\frac{\sin 2^{n} \theta}{2^{n} \sin \theta}$
Sol :
P(n): $\cos \theta \cdot \cos 2 \theta \cdot \cos 4 \theta \ldots \cos 2^{n-1} \theta=\frac{\sin 2^{n} \theta}{2^{n} \sin \theta}$

Step-1
When n=1

L.H.S $=\cos 2^{1-1} \theta=\cos 2^{0} \theta=\cos \theta$

R.H.S$=\frac{\sin 2 \theta}{2 \sin \theta}=\frac{2 \sin \theta \cos \theta}{-2 \sin \theta}=\cos \theta$

L.H.S=R.H.S

P(1) is true

Step-2
Let P(k) be true , then
$P(k): \cos \theta \cdot \cos 2 \theta \cdot \cos 4 \theta \ldots \cos 2^{k-1} \theta =\frac{\sin 2^{k} \theta}{2^{k} \sin \theta}$

prove that P(k+1) be true

$P(k+1): \cos \theta \cdot \cos 2\theta \cdot \cos 4 \theta \ldots \ldots \cos 2^{k-1}\cos 2^{k} \theta=\frac{\sin 2^{k+1} \theta}{2^{k+1} \sin \theta}$..(ii)

multiplying both sides by $\cos 2^{k} \theta$ in (i)

$\cos \theta \cdot \cos 2\theta \cdot \cos 4 \theta-\ldots \cos 2^{k-1} \theta \cdot \cos 2^{k} \theta=\frac{\sin 2^{k} \theta}{2^{k} \sin \theta} \times \cos 2^{k} \theta$

$=\frac{2 \sin 2^{k} \theta \cdot \cos ^{k} \theta}{2 \cdot 2^{k} \sin \theta}$

$=\frac{\sin 2^{1} \cdot 2^{k} \theta}{2^{k+1} \sin \theta}$

$=\frac{\sin 2^{k+1} \theta}{2^{k+1} \sin \theta}$

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 24

$\sin \alpha+\sin 2 \alpha+\ldots+\sin n \alpha=\frac{\sin \frac{n \alpha}{2}}{\sin \frac{\alpha}{2}} \sin \frac{(n+1) \alpha}{2}$
Sol :
P(n): $\sin \alpha+\sin 2 \alpha+\ldots+\sin n \alpha=\frac{\sin \frac{n \alpha}{2}}{\sin \frac{\alpha}{2}} \sin \frac{(n+1) \alpha}{2}$

Step-1
when n=1

L.H.S=sin(1)𝛼=sin𝛼

R.H.S$=\frac{\sin \frac{1 \times \alpha}{2}}{\sin \frac{\alpha}{2}} \cdot \sin \frac{(1+1) \alpha}{2}$

$=\frac{\sin \frac{\alpha}{2}}{\sin \frac{\alpha}{2}} \times \sin \frac{2\alpha}{2}=sin \alpha$

L.H.S=R.H.S

P(1) is true

Step-2
Let P(k) be true
$P(k): \sin \alpha+\sin 2 \alpha+\ldots+\sin k \alpha=\frac{\sin \frac{k \alpha}{2}}{\sin \frac{\alpha}{2}} \sin \frac{(k+1)}{2}\alpha$..(i)

then prove that P(k+1) is true

P(k+1): $\sin \alpha+\sin 2 \alpha+\ldots+\sin k \alpha+\sin (k+1) \alpha=\dfrac{\frac{\sin (k+1) \alpha}{2}}{\sin \frac{\alpha}{2}} \cdot \sin \frac{(k+2) \alpha}{2}$..(ii)

adding both sides by sin(k+1)𝛼 in (i)

$\sin x+\sin 2 \alpha+\ldots-+\sin k \alpha+\sin (k+1) \alpha=\frac{\sin \frac{k \alpha}{2}}{\sin \frac{\alpha}{2}} \sin \frac{(k+1)}{2} \alpha +\sin(k+1)\alpha$

$=\frac{\sin \frac{k \alpha}{2}}{\sin \frac{\alpha}{2}} \sin \frac{(k+1) \alpha}{2}+2 \sin \frac{(k+1)}{2} \alpha \cos \left(\frac{k+1}{2}\right) \alpha$

$=\sin \frac{(k+1) \alpha}{2}\left[\frac{\sin \frac{k \alpha}{2}}{\sin \frac{\alpha}{2}}+2 \cos \left(\frac{k+1}{2}\right) \alpha\right]$

$=\frac{\sin (k+1) \alpha}{2}\left[\frac{\sin \frac{k \alpha}{2}+2 \cos \left(\frac{k+1}{2}\right)-\sin \frac{\alpha}{2}}{\sin \frac{\alpha}{2}}\right]$

$=\dfrac{\frac{\sin (k+1) \alpha}{2}}{\sin \frac{\alpha}{2}}\left[\sin \frac{k \alpha}{2}+\sin \left(\frac{(k+1) \alpha}{2}+\frac{\alpha}{2}\right)-\sin\left(\frac{(k+1)\alpha}{2}-\frac{\alpha}{2}\right)\right]$

$=\dfrac{\frac{\sin (k+1)}{2}}{\sin \frac{\alpha}{2}}\left[\sin \frac{k x}{2}+\sin \frac{(k+2) x}{2}-\sin \frac{k}{2}\right]$

$=\frac{\sin \frac{(k+1) \alpha}{2}}{\sin \frac{\alpha}{2}} \cdot \sin \frac{(k+2)\alpha}{2}$

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 25

$\sin \frac{\pi}{3}+\sin \frac{2 \pi}{3}+\ldots+\sin \frac{n \pi}{3}=2 \sin \frac{n \pi}{6} \sin \frac{(n+1) \pi}{6}$
Sol :
P(n): $\sin \frac{\pi}{3}+\sin \frac{2 \pi}{3}+\ldots+\sin \frac{n \pi}{3}=2 \sin \frac{n \pi}{6} \sin \frac{(n+1) \pi}{6}$

Step-1
when n=1

L.H.S$=\sin \frac{1 \times \pi}{3}=\frac{\sqrt{3}}{2}$

R.H.S$=2 \sin \frac{1\times \pi}{6} \sin \frac{(1+1) \pi}{6}$

$=2 \times \frac{1}{2} \sin \frac{2 \pi}{-6}=\frac{\sqrt{3}}{2}$

P(1) is true

Step-2
Let P(k) be true, then
$P(k): \sin\frac{\pi}{3}+\sin \frac{2 \pi}{3}+\ldots\cdots+\sin \frac{k \pi}{3}=2 \sin \frac{k \pi}{6} \sin \frac{(k+1)}{6}\pi$..(i)

then prove that P(k+1) is true
$P(k+1): \sin \frac{\pi}{3}+\sin \frac{2 \pi}{3}+\ldots \ldots+\sin \left(\frac{k+1}{3}\right) \pi=2 \sin \frac{(k+1) \pi}{6} \sin \frac{(k+2) \pi}{6}$..(ii)

adding both sides $\sin \left(\frac{k+1}{3}\right) \pi$ in (i)

$\sin \frac{\pi}{3}+\sin \frac{2 \pi}{3} +--+\sin \frac{k \pi}{3}+\frac{\sin (k+1) \pi}{3}=2 \sin \frac{k \pi}{6} \sin \frac{(k+1) \pi}{6}+\sin \frac{(k+1) \pi}{3}$

$=2 \sin \frac{k \pi}{6} \sin \frac{(k+1) \pi}{6}+2 \sin \frac{(k+1) \pi}{6} \cos \frac{(k+1) \pi}{6}$

$=2 \sin \frac{(k+1) \pi}{6}\left[\sin \frac{k \pi}{6}+\cos \frac{(k-1) \pi}{6}\right]$

$=2 \sin \frac{(k+1) \pi}{6}\left[\sin \frac{k \pi}{6}+\sin \left[\frac{\pi}{2}-\left(\frac{k \pi}{6}+\frac{\pi}{6}\right)\right]\right.$

$=2 \sin \frac{(k+1) \pi}{6}\left[\sin \frac{k \pi}{6}+\sin \left(\frac{\pi}{3}-\frac{k\pi }{6}\right)\right]$

$=2 \sin \frac{(k+1)\pi} {6}\left[2 \times \sin \frac{\frac{k \pi}{{6}}+\frac{\pi}{3}-\frac{k \pi}{6}}{2} \cos \frac{\frac{k \pi}{6}-\frac{\pi}{3}+\frac{k \pi}{6}}{2}\right]$

$=2 \sin \frac{(k+1) \pi}{6}\left[2 \sin \frac{\pi}{6} \cdot \cos \dfrac{\frac{2 k \pi}{{6}}-\frac{\pi}{3}}{2}\right]$

$=2 \sin \frac{(k+1) \pi}{6}\left[2 \times \frac{1}{2}+\cos \left(\frac{k \pi}{6}-\frac{\pi}{6}\right)\right]$

$=2 \sin \frac{(k+1) \pi}{6} \times \cos \left[-\left(\frac{\pi}{6}-\frac{k \pi}{6}\right)\right]$

$=2 \sin \frac{(k+1) \pi}{6} \cos \left(\frac{\pi}{6}-\frac{k \pi}{6}\right)$

$=2 \sin \frac{(k+1) \pi}{6} \sin \left[\frac{\pi}{2}-\frac{\pi}{6}+\frac{k \pi}{6}\right]$

$=2 \sin \frac{(k+1) \pi}{6} \sin \left(\frac{k \pi}{6}+\frac{\pi}{3}\right]$

$=2 \sin \frac{(k+1) \pi}{6} \sin \left(\frac{k \pi+2 \pi}{6}\right)$

$=2 \sin \frac{(k+1) \pi}{6} \sin \frac{(k+2)\pi}{6}$

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 27

$(1+i)^{n}=2^{\frac{n}{2}}\left(\cos \frac{n \pi}{4}+i \sin \frac{n \pi}{4}\right)$
Sol :
Let

P(n): $(1+i)^{n}=2^{\frac{n}{2}}\left(\cos \frac{n \pi}{4}+i \sin \frac{n \pi}{4}\right)$

Step-1
when n=1

L.H.S=(1+i)¹=1+i

R.H.S=$=2^{1 / 2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)$

$=\sqrt{2}\left(\frac{1}{\sqrt{2}}+i \times \frac{1}{\sqrt{2}}\right)$

$=\sqrt{2} \times \frac{1}{\sqrt{2}}(1+i)$

=1+i

L.H.S=R.H.S

P(1) is true

Step-2
Let P(k) be true , then
P(k):$(1+i)^{k}=2^{k / 2}\left(\cos \frac{k \pi}{4}+i \sin \frac{k \pi}{4}\right)$..(i)

then prove that P(k+1) is true

P(k+1): $(1+i)^{k+1}=2^{k+1}\left[\cos \frac{(k+1) \pi}{4}+i \frac{\sin (k+1) \pi}{4}\right]$..(ii)

multiplying both bides by (1+i) in (i)

$(1+i)^{k}(1+i)^{1}=2^{k / 2}\left(\cos \frac{k \pi}{4}+i \sin \frac{k \pi}{4}\right)(1+i)$

$(1+i)^{k+1}=2^{k / 2}\left[\left(\cos \frac{k \pi}{4}+i \sin \frac{k\pi}{4}\right) \sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)\right.$

$=2^{k/{2}} \cdot 2^{1 / 2}\left[\cos \frac{ k \pi}{4} \cos \frac{\pi}{4}+i \cos \frac{k \pi}{4} \sin \frac{\pi}{4}+i \sin \frac{k \pi}{4} \cos \frac{\pi}{4}+i^{2} \sin \frac{k \pi}{4} \sin \frac{\pi}{4}\right]$

$=2^{\frac{k+1}{2}}\left[\left(\cos \frac{\pi}{4} \cos \frac{\pi}{4}-\sin\frac{k\pi}{4} \sin \frac{\pi}{4}\right)+\left.i\left(\sin \frac{k \pi}{4} \cos \frac{\pi}{4}+\cos \frac{k \pi}{4}\sin \frac{\pi}{4}\right)\right]\right.$

$=2^{\frac{k-1}{2}}\left[\cos \left(\frac{k \pi}{4}+\frac{\pi}{4}\right)+i \sin \left(\frac{k\pi}{4}+\frac{\pi}{4}\right)\right]$

$=2^{\frac{k+1}{2}}\left[\cos \left(\frac{k\pi+\pi}{4}\right)+i \sin \left(\frac{k \pi+\pi}{4}\right)\right]$

$=2^{\frac{k+1}{2}}\left[\cos \frac{(k+1) \pi}{4}+i \sin \frac{(k+1)\pi}{4} \right]$

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 28

Sol :
$P(n):(a b)^{n}=a^{n} b^{n}$

Step-1
when n=1

L.H.S=(ab)¹=ab

R.H.S=a¹b¹=ab

L.H.S=R.H.S

P(1) is true

Step-2
Let P(k) be true , then

$P(k):( a b)^{k}=a^{k} b^{k}$..(i)

prove that P(k+1) is true

$P(k+1): (a b)^{k+1}=a^{k+1} b^{k+1}$..(ii)

multiplying both sides by ab in (i)

$(a b)^{k} \cdot(a-b)^{1}=a^{k} b^{k} \cdot a^{1} b^{1}$

$(a b)^{k+1}=a^{k+1} b^{k+1}$

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 29

$a_{n+1}=\frac{1}{(n+1) !}$
Sol :
Let P(n): $a_{n+1}=\frac{1}{(n+1) !}$

Step-1

when n=1

L.H.S=$=a_{1+1}=a_{2}=\frac{a_{1}}{1+1}=\frac{1}{2}$

R.H.S$=\frac{1}{(1+1) !}=\frac{1}{2!}=\frac{1}{2 \times 1}=\frac{1}{2}$

L.H.S=R.H.S

P(1) is true

Step-2

Let P(k) be true , then

$P(k): a_{k+1}=\frac{1}{(k+1)!}$..(i)

prove that P(k+1) is true 

$P(k+1): \quad a_{k+2}=\frac{1}{(k+2) !}$..(ii)

$a_{n+1}=\frac{a_{n}}{n+1}$ [given]

or 

n=k , $a_{k+1}=\frac{a_{k}}{k+1}$

$a_{k+2}=\frac{a_{k+1}}{k+2}$

$=\frac{1}{k !(k+1)(k+2)}$

$a_{k+2}=\frac{1}{(k+2) !}$

Question 30

Sol :
Let $7^{n}-3^{n}$ is divisible by 4

Step-1
when n=1

P(1)=$7^{\prime}-3^{\prime}=4$  , divisible by 4

P(1) is true

Step-2
Let P(k) be true , then

$P(k): 7^{k}-3^{k}$ , divisible by 4

or $7^{k}-3^{k}=4 \lambda$

$\Rightarrow 7^{k}=4 \lambda+3^{k}$

prove that P(k+1) is true

$7^{k+1}-3^{k+1}$

$=7^{1}-7^{k}-3^{1} \cdot 3^{k}$

$=7.7^{k}-3 \cdot 3^{k}$

$=7\left(4 \lambda+3^{k}\right)-3 \cdot 3^{k}$

$=28 \lambda+7 \cdot 3^{k}-3 \cdot 3^{k}$

$=28 \lambda+4 \cdot 3^{k}$

$=4\left(7 \lambda+3^{k}\right)$

∴ $7^{k+1}-3^{k+1}$ is divisible by 4

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 32

Sol :
Let $n(n+1)(n+5)$ , is divisible by 3

Step-1
when n=1

P(1)=1(1+1)(1+5)
=1(2)(6)
=12 which is divisible by 3

P(1) is true

Step-2
Let P(k) be true , then

$P(k): k(k+1)(k+5)$ , which is divisible by 3

$P(k): k(k+1)(k+5)=3 \lambda$

prove that (k+1) is true

$P(k+1):(k+1)(k+2)(k+6)$ , divisible by 3

(k+1)(k+2)(k+6)

=(k+1)(k+2)[(k+5)+1]

=(k+1)(k+2)(k+5)+(k+1)(k+2)

=k(k+1)(k+5)+2(k+1)(k+5)+(k+1)(k+2)

=3𝜆+(k+1)[2(k+5)+k+2]

=3𝜆+(k+1)(2k+10+k+2)

=3𝜆+(k+1)(3k+12)

=3𝜆+(k+1)(k+4)

=3[𝜆+(k+1)(k+4)]

divisible by 3

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 35

$4^{n}-3 n-1$ , is divisible by 9
Sol :
Let
P(n): $4^{n}-3 n-1$ is divisible by 9

Step-1
when n=1

$P(1): 4^{1}-3(1)-1=4-3-1=0$ , divisible by 9

P(1) is true

Step-2
Let P(k) be true , then

$P(k): 4^{k}-3 k-1$ , divisible by 9

$4^{k}-3 k-1=9 \lambda$

$4^k=9 \lambda+3 k+1$

Prove that P(k+1) is true

$p(k+1): 4^{k+1}-3(k+1)-1$,  divisible by 9
$4^{k+1}-3(k+1)-1$
$=4^{k} \times 4-3 k-3-1$
$=(9 \lambda+3k+1) 4-3 k-4$
=36𝜆+12k+4-3k-4
=36𝜆+9k
=9(4𝜆+k)
divisible by 9

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 37

Sol :
$5 \cdot 2^{3 n-2}+3^{3 n-1}$ is divisible by 19

Step-1
when n=1

$P(1): 5 \cdot 2^{3(1)-2}+3^{3(1)-1}$

$=5 \cdot 2^{1}+3^{2}$
=10+9
=19 , 19 is divisible by 19

P(1) is true

Step-2
Let P(k)  be true , then
$p(k): 5 \cdot 2^{3 k-2}+3^{3 k-1}$ , divisible by 19

$5 \cdot 2^{3 k-2}+3^{3 k-1}=19\lambda$

Prove that P(k+1) is true

$P(k+1): 5 \cdot 2^{3 k+1}+3^{3 k+2}$ , divisible by 19

$=15 \cdot 2^{3 k-2+3}+3^{3 k+2}$

$=15 \cdot 2^{3 k-2} \times 2^{3}+3^{3 k+2}$

$=\left(19 \lambda-3^{3 k-1}\right) \times 8+3^{3 k+2}$

$=152 \lambda-8 \cdot 3^{3 k-1}+3^{3 k+2}$

$=152 \lambda-8 \frac{3^{3 k}}{3}+3^{3 k} \times 3^{2}$

$=152 \lambda-\frac{8}{3} \cdot 3^{3 k}+9 \cdot 3^{3 k}$

$=152\lambda \cdot \frac{-8 \cdot 3^{3 k}+27 \cdot 3^{3k}}{3}$

$=152 \lambda+\frac{19}{3} \cdot 3^{3 k}$

$=152 \lambda+19.3^{3 k-1}$

$=19\left(8 \lambda+3^{3 k-1}\right)$

divisible by 19

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 43

$\frac{n^{3}}{3}+n^{2}+\frac{5}{3} n+1$
Sol :
Let
P(n): $\frac{n^{3}}{3}+n^{2}+\frac{5}{3} n+1$ is a natural number

Step-1
when n=1

$P(1)=\frac{1^{3}}{3}+1^{2}+\frac{5}{3}(1)+1$

$=\frac{1}{3}+1+\frac{5}{3}+1$

$=\frac{1+3+5+3}{3}=\frac{12}{13}=4$

4 is a natural number

P(1) is true

Step-2
Let p(k) be true , then

$P(k): \frac{k^{3}}{3}+k^{2}+\frac{5}{3} k+1$ [is a natural number]

$\frac{k^{3}}{3}+k^{2}+\frac{5}{3} k+1=\lambda,\lambda \in N$

Prove that P(k+1) is true

$P(k+1): \frac{(k+1)^{3}}{3}+(k+1)^{2}+\frac{5}{3}(k+1)+1$ [is a natural number]

$\frac{(k+1)^{3}}{3}+(k+1)^{2}+\frac{5}{3}(k+1)+1$

$=\frac{1}{3}\left[k^{3}+3 \cdot k^{2} \cdot 1+3 k \cdot 1^{2}+1^{3}\right]+\left(k^{2}+2 \cdot k \cdot 1+1^{2}\right)+\frac{5}{3}k+\frac{5}{3}+1$

$=\frac{1}{3}\left[k^{3}+3 k^{2}+3 k+1\right]+k^{2}+2 k+1+\frac{5}{3} k+\frac{5}{3}+1$

$=\frac{1}{3} k^{3}+k^{2}+k+\frac{1}{3}+k^{2}+2 k+1+\frac{5}{3} k+\frac{5}{3}+1$

$=\left(\frac{1}{3} k^{3}+k^{2}+\frac{5}{3} k+1\right)+k^{2}+3 k+\frac{1}{3}+1+\frac{5}{3}$

$=\lambda+k^{2}+3 k+\frac{1+3+5}{3}$

$=\lambda+k^{2}+3 k+\frac{9}{3}$

$=\lambda+k^{2}+3 k+3$  is a natural number

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 44

$x^{n}+y^{n}$ is divisible by x+y
Sol :
Let
P(n): $x^{n}+y^{n}$ is divisible by x+y

Step-1
when n=1

P(1)=$x^{\prime}+y^{\prime}=x+y$ , divisible by x+y

P(1) is true

Step-2
Let P(k) be true , then

$p(x): x^{k}+y^{k}, \quad x+y$ is divisible by x+y

$x^{k}+y^{k}=\lambda(x+y)$

Prove that P(2k+1) is true

$P(2 k+1): x^{2 k+1}+y^{2 k+1}$ , is divisible by x+y

$x^{2 k+1}+y^{2 k+1}$

$=x \cdot x^{2 k}+y \cdot y^{2 k}+x^{2 k} y-x^{2 k} +y^{2 k} \cdot x-y^{2 k} \cdot x$

$=x^{2 k}(x+y)+y^{2 k}(x+y)-x y\left(z^{2 k-1}-y^{2 k-1}\right)$

P(2k+1): is divisible by x+y

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 45

$x\left(x^{n-1}-n a^{n-1}\right)+a^{n}(n-1)$ divisible by $(x-a)^{2}$

Sol :
Let 

P(n) :$x\left(x^{n-1}-n a^{n-1}\right)+a^{n}(n-1)$ divisible by $(x-a)^{2}$

Step-1

when n=1

$p(1): x\left(x^{(-1}-1 \cdot a^{1-1}\right)+a^{1}(1-1)$

=x(1-1)+a(0)

=0

divisible by $(x-a)^{2}$ 

P(1) is true

Step-2

Let P(k) be true , then

$P(k):  x\left(x^{k-1}-k a^{k-1}\right)+a^{k}(k-1)$ divisible by $(x-a)^{2}$ divisible by $(x-a)^{2}$

$x\left(x^{k-1}-k a^{k-1}\right)+a^{k}(k-1)=\lambda(x-a)^2$

Prove that P(k+1) is true 

$P(x+1): \quad x\left[x^{k}-(k+1) a^{k}\right]+a^{k+1} \cdot k$ , divisible by $(x-a)^{2}$

[]

(b)

(i) $\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7 n}{15}$ is a natural number

Sol :

Let $\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7 n}{15}$ is a natural number

Step-1

when n=1

$P(1): \frac{1^{5}}{5}+\frac{1^{3}}{3}+\frac{7}{15}(1)$

$=\frac{1}{5}+\frac{1}{3}+\frac{7}{15}$

$=\frac{3+5+7}{15}=\frac{15}{15}=1$

P(1) is true

Step-2

Let P(k) be true , then

$P(k)=\frac{k^{5}}{5}+\frac{k^{3}}{3}+\frac{7}{15} k$ is a natural number

$\frac{x^{5}}{5}+\frac{k^{3}}{3}+\frac{7}{15} k=\lambda, \quad \lambda \in N$

Prove that P(k+1) is true 

$P(k+1): \frac{(k+1)^{5}}{5}+\frac{(k+1)^{3}}{3}+\frac{7}{15}(k+1)$ is a natural number

$\frac{(k+1)^{5}}{5}+\frac{(k+1)^{3}}{3}+\frac{7}{15}(k+1)$

$=\frac{1}{5}\left[k^{5}+5 k^{4}+10 k^{3}+10 k^{2}+5 k+1\right]+\frac{1}{3}\left[k^{3}+3 k^{2}+3 k+1\right]+\frac{7}{15}(k+1)$

$=\frac{1}{5} k^{5}+k^{4}+2 k^{3}+2 k^{2}+k+\frac{1}{5}+\frac{1}{3} k^{3}+k^{2}+k+\frac{1}{3}+\frac{7}{15}k+\frac{7}{15}$

$=\left(\frac{1}{5} k^{5}+\frac{1}{3} k^{3}+\frac{7}{15} k\right)+k^{4}+2 k^{3}+3 k^{2}+2 k+\frac{1}{3}+\frac{1}{5}+\frac{7}{15}$

$=\lambda+k^{4}+2 k^{3}+3 k^{2}+2 k+\frac{5+3+7}{15}$

$=x+k^{4}+2 k^{3}+3 k^{2}+2 k+1$

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 46

(i) $2^{n}>n$

Sol :

Let

P(n):$2^{n}>n$

Step-1

when n=1

$2^{1}>1 \Rightarrow 2>1$

P(1) is true

Step-2

Let P(k) be true , then

$2^{k}>k$..(i)

Prove that P(k+1) is true

$2^{k+1}>(k+1)$

Multiplying with 2 both sides in (i)

$2^{k} \cdot 2>2 k$

$2^{k+1}>k+k$

$2^{k+1}>(k+1)$

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 48

$\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}>\frac{13}{24}, n>1$

Sol :

Let $\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}>\frac{13}{24}, n>1$

Step-1

when n=2

$P(2): \frac{1}{2+1}+\frac{1}{2+2}+--\quad+\frac{1}{2(2)}>\frac{13}{24}$

P(2) is true

Step-2

Let P(k) be true , then

$P(k): \frac{1}{k+1}+\frac{1}{k+2}+--\frac{1}{2 k}>\frac{13}{24}, k>1$

Prove P(k+1) is true

$P(k+1): \frac{1}{k+2}+\frac{1}{k+5}+-\cdots+\frac{1}{2(k+1)}>\frac{13}{24}$

$\frac{1}{k+2}+\frac{1}{k+3}+---\frac{1}{2(k+1)}=\left(\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+---\frac{1}{2 k}\right)+\frac{1}{2 k-1}+\frac{1}{2 k+2}-\frac{1}{k+1}$

$=\left(\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+\ldots-\frac{1}{2 k}\right)+\frac{1}{2k+1}+\frac{1}{2(k+1)}-\frac{1}{k+1}$

$=\left(\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+\cdots-\frac{1}{2 k}\right)+\frac{1}{2k+1}-\frac{1}{2(k+1)}>\frac{13}{24}$

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.